Question 42 of 58intermediate🔧 ApplyLong Answer4 marks

Solve system of linear equations, using matrix method: 4x - 3y = 3 3x - 5y = 7

Correct Answer

x = -6/11 and y = -19/11

Exercise: EXERCISE 4.5 | Q: 9 | (Page 22)
For More Understanding

Explanation

The question requires solving a system of linear equations using the matrix method. Following the methodology demonstrated in Example 17 of the provided context, the system is converted into the matrix form AX = B. The solution is found by calculating the determinant to ensure the inverse exists, finding the adjoint of the coefficient matrix, and multiplying the inverse by the constant matrix B to find X.

Solution Steps

  1. Step 1: Write the system in matrix form The system of equations 4x−3y=34x - 3y = 3 and 3x−5y=73x - 5y = 7 can be written as AX=BAX = B, where:

    A=[4−33−5],X=[xy], and B=[37]A = \begin{bmatrix} 4 & -3 \\ 3 & -5 \end{bmatrix}, X = \begin{bmatrix} x \\ y \end{bmatrix}, \text{ and } B = \begin{bmatrix} 3 \\ 7 \end{bmatrix}

  2. Step 2: Calculate the determinant of A

    ∣A∣=(4)(−5)−(−3)(3)=−20+9=−11|A| = (4)(-5) - (-3)(3) = -20 + 9 = -11

    Since ∣A∣≠0|A| \neq 0, AA is nonsingular and its inverse exists.

  3. Step 3: Find the adjoint of A The cofactors are A11=−5A_{11} = -5, A12=−3A_{12} = -3, A21=3A_{21} = 3, and A22=4A_{22} = 4. Thus, adj(A)=[−53−34]\text{adj}(A) = \begin{bmatrix} -5 & 3 \\ -3 & 4 \end{bmatrix}.

  4. Step 4: Calculate the inverse A−1A^{-1}

    A−1=1∣A∣adj(A)=−111[−53−34]A^{-1} = \frac{1}{|A|} \text{adj}(A) = -\frac{1}{11} \begin{bmatrix} -5 & 3 \\ -3 & 4 \end{bmatrix}

  5. Step 5: Solve for X

    X=A−1B=−111[−53−34][37]X = A^{-1}B = -\frac{1}{11} \begin{bmatrix} -5 & 3 \\ -3 & 4 \end{bmatrix} \begin{bmatrix} 3 \\ 7 \end{bmatrix} X=−111[−15+21−9+28]=−111[619]=[−6/11−19/11]X = -\frac{1}{11} \begin{bmatrix} -15 + 21 \\ -9 + 28 \end{bmatrix} = -\frac{1}{11} \begin{bmatrix} 6 \\ 19 \end{bmatrix} = \begin{bmatrix} -6/11 \\ -19/11 \end{bmatrix}

    Hence, x=−6/11x = -6/11 and y=−19/11y = -19/11.

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